已知等差数列{an}中,a2=5,a5=9.
问题描述：
已知等差数列{an}中,a2=5,a5=9.
(1)求等差数列{an}的通项公式.【已求：an=11\3+4\3 (n-1）】
（2）设bn=2^an,求该数列{bn}的前n项和 Sn.
答
{an}公差d = (a5 - a2 )/3 = (9-5)/3 = 4/3
a1 = a2- 4/3 = 5- 4/3= 11/3
an=11/3+4/3 (n-1）
bn = 2^an
bn/bn-1 = 2^an/2^an-1 = 2^(an-an-1) = 2^d = 2^(4/3)
bn是等比数列
b1 = 2^a1 = 2^(11/3)
q = 2^(4/3)
{bn}的前n项和
Sn= b1* (1-q^n)/(1-q)= 2^(11/3) *(1-(2^(4/3))^n)/ (1-2^(4/3))
= 2^(11/3) *(2^(4n/3) -1)/((2^(4/3)-1)